Put
\(u_k(x)=r_k^{\frac{n-2}{2}}u(r_kx)c_k,\) 
where \(\int _{\partial B_1}\log u_k d{\mathbb {S}^{n-1}}=0\) . It is easy to check that
\(-\Delta _{g_k} u_k=f_ku_k\) , where
\(f_k=-c(n)r_k^2R_{g_0}(r_kx)+c(n)R(r_kx)(r_k^\frac{n-2}{2}u_k(r_kx))^\frac{4}{n-2},\) 
which satisfies, for sufficiently large \(k\) ,
\(\Vert f_k\Vert _{L^\frac{n}{2}(B_r\setminus B_\frac{1}{r})}\le Cr_k^2+C\left(\int _{B_{rr_k}\setminus B_\frac{r_k}{r}}|R|^\frac{n}{2}u^\frac{2}{n}dx\right)^\frac{n-2}{2n}\le \min \lbrace \epsilon _0,\epsilon _0^{\prime }\rbrace .\) 
By Lemma REF  and the Poincaré inequality ( c.f. [1]}),
\(\log u_k\)  is bounded in \(W^{1,2}(B_r\setminus B_\frac{1}{r})\) .
Then, by Lemma REF , \(u_k\)  is bounded in \(W^{2,p}(B_r\setminus B_\frac{1}{r})\) , which follows that \(u_k\)  converges weakly to a positive harmonic function \(u^{\prime }\) 
locally on \(\mathbb {R}^n\setminus \lbrace 0\rbrace \)  with \(\int _{\partial B_1}\log u^{\prime }d\mathbb {S}^{n-1}=0\) .
By Corollary 3.14 in [2]}, \(u^{\prime }\)  is written as
\(u^{\prime }=a+b|x|^{2-n},\) 
where \(a\)  and \(b\)  are nonnegative constants.
Applying Theorem REF , we get
\(\int _{B_{r_kr}\setminus B_{r_krL^{-1}}}\frac{u^2}{|x|^2}dV_{g_0}\le \frac{1}{L}\int _{B_{r_krL^{-1}}\setminus B_{r_krL^{-2}}}\frac{u^2}{|x|^2}dV_{g_0},\) 
which means that
\(\int _{B_r\setminus B_{rL^{-1}}}\frac{u_k^2}{|x|^2}dV_{g_{0,k}}\le \frac{1}{L}\int _{B_{rL^{-1}}\setminus B_{rL^{-2}}}\frac{u_k^2}{|x|^2}dV_{g_{k}},\) 
where \(g_{k,ij}(x)=g_{0,ij}(r_kx)\) .
Taking the limit, we obtain
\(\int _{B_r\setminus B_{rL^{-1}}}\frac{{u^{\prime }}^2}{|x|^2}dx\le \frac{1}{L}\int _{B_{rL^{-1}}\setminus B_{rL^{-2}}}\frac{{u^{\prime }}^2}{|x|^2}dx.\) 
Letting \(r\)  be sufficiently large in the above inequality, we get \(a=0\) .
Since \(\int _{\partial B_1}\log u^{\prime }=0\) , \(b=1\) .
If we change coordinates: \(x\rightarrow \frac{x}{|x|^2}\) , then \({u^{\prime }}^\frac{4}{n-2}g_{euc}\) 
is just \(g_{euc}\)  in the new coordinates. The convergence of \(d_{g_k}\)  follows
from Proposition REF  directly.
\(\hfill \Box \)
In the above lemma, we fail to write \(c_k\)  in terms of \(r_k\) . Hence the behaviors of
\(u(r_kx)\)  are not really understood. However, the lemma is already strong enough
to help us get the following decays:
Corollary 3.2 
For any \(\tau \in (0,1)\) , there exists \(\delta \) , such that
for any \(r<\delta \) , the following hold
\((1-\tau )\frac{1}{2^n}\le \frac{\mathrm {vol}(B_{2r}\setminus B_r,g)}{\mathrm {vol}(B_r\setminus B_{r/2},g)}\le \frac{1}{2^n}(1+\tau );\\(1-\tau )(2^n-1)\le \frac{\mathrm {vol}(B_{2r}\setminus B_r)}{V_n (d_g(2rx_0,rx_0))^n}\le (2^n-1)(1+\tau ),\,\,\,\,\forall x_0\in \partial B;\\\frac{\int _{B_{r}\setminus B_{r/2}}|x|^\beta |\nabla _{g_0} u|^\alpha dV_{g_0}}{\int _{B_{2r}\setminus B_{r}} |x|^\beta |\nabla _{g_0} u|^\alpha dV_{g_0}}<2^{(n-1)\alpha -n-\beta }(1+\tau ),\,\,\,\,\forall \alpha \in [1,n), \,\,\,\,\beta \in \mathbb {R};\\\frac{\int _{B_{r}\setminus B_{r/2}}|x|^\beta u^\alpha dV_{g_0}}{\int _{B_{2r}\setminus B_{r}} |x|^\beta u^\alpha dV_{g_0}}<2^{(n-2)\alpha -n-\beta }(1+\tau ),\,\,\,\,\forall \alpha \in [1,+\infty ),\,\,\,\,\beta \in \mathbb {R};\\\frac{1}{2^{n-2}}(1-\tau )\le \frac{\int _{\partial B_r}|\nabla _{g_0}\log u|dS_{g_0}}{\int _{\partial B_{2r}}|\nabla _{g_0}\log u|dS_{g_0}}\le \frac{1}{2^{n-2}}(1+\tau );\\2^{-1}(1-\tau )\le \frac{d_g(2rx_0,rx_0)}{d_g(rx_0,rx_0/2)}\le 2^{-1}(1+\tau ),\,\,\,\,\forall x_0\in \partial B;\\(1-\tau )\le \frac{d_g(rx_0,\partial B_{2r})}{d_g(rx_0,2rx_0)}\le (1+\tau ),\,\,\,\,\forall x_0\in \partial B;\\2^{-1}(1-\tau )\le \frac{diam(\partial B_{2r})}{diam(\partial B_r)}\le 2^{-1}(1+\tau ).\) 

Let \(u_k=c_kr_k^\frac{n-2}{2}u_k(r_kx)\)  be as in the proof of Lemma REF , which converges to \(u^{\prime }=|x|^{2-n}\) 
weakly in \(W^{2,p}_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\)  for any \(p<\frac{n}{2}\) .
Then we may assume \(\nabla u_k\)  converges in \(L^q_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\) 
for any \(q<n\) , and \(u_k\)  converges in \(L^q_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\)  for any
\(q\) . By the trace inequality, we can also assume
\(\log u_k\)  converges in \(L^1(\partial B_t)\) .
Now, we prove (REF ):
assume it is not true, then there exists \(r_k\rightarrow 0\) , such that
\(\frac{\mathrm {vol}(B_{2r_k}\setminus B_{r_k},g)}{\mathrm {vol}(B_{r_k}\setminus B_{r_k/2},g)}> \frac{1}{2^n}(1+\tau ),\) 
which means that
\(\frac{\mathrm {vol}(B_{2}\setminus B_1,g_k)}{\mathrm {vol}(B_{1}\setminus B_{1/2},g_k)}> \frac{1}{2^n}(1+\tau ),\) 
where \(g_{k,ij}=u_k^\frac{4}{n-2}g_{0,ij}(r_kx)\) . Letting \(k\rightarrow +\infty \) ,
we get
\(\frac{\mathrm {vol}(B_{2}\setminus B_1,g_\infty )}{\mathrm {vol}(B_{1}\setminus B_{1/2},g_\infty )}\ge \frac{1}{2^n}(1+\tau ),\) 
where \(g_\infty =|x|^{2-n}g_{euc}\) . A contradiction.
Since the proofs to other inequalities are almost the same, we omit them.
\(\hfill \Box \)
The inequalities ()-() will be used to estimate quantities on \(B_{2r}\setminus B_r\) . For example, by (REF ),
\(\mathrm {vol}(B_{2^kr}\setminus B_{2^{k-1}r},g)\le \left(\frac{1+\tau }{2^n}\right)^k\mathrm {vol}(B_{2r}\setminus B_r,g),\) 
then
\(\mathrm {vol}(B\setminus B_r,g)\le C\mathrm {vol}(B_{2r}\setminus B_{r},g).\) 
We give several more applications.
Corollary 3.3 
\((\overline{B}\setminus \lbrace 0\rbrace ,g)\)  is complete.

Let \(\lbrace x_k\rbrace \)  be a sequence which does not contain a convergent subsequence. Then
\(x_k\rightarrow 0\) . Fix an \(x_0\in \partial B\) . For our purposes, it suffices to show
\(d_g(x_k,x_0)\rightarrow +\infty \) .
By () and (), \(d_g(x_k,2x_k)\rightarrow +\infty \) ,
and
\(d_g(x_0,x_k)\ge d_g(x_k,\partial B_{2|x_k|})\ge (1-\tau )d_g(x_k,2x_k)\rightarrow +\infty .\) 
\(\hfill \Box \)
Corollary 3.4 
We have
\(\lim _{r\rightarrow +\infty }\frac{\mathrm {vol}(B_r^g(x_0),g)}{V_nr^n}=1\) 
and \((\overline{B}\setminus \lbrace 0\rbrace ,r^{-2}g,x_0)\) 
converges to \(\mathbb {R}^n\)  in the Gromov-Hausdorff distance
for any \(x_0\) .

Since the proofs for volume convergence and Gromov-Hausdorff
convergence are almost the same, we only prove the volume convergence
here.
Let \(\rho _k\rightarrow +\infty \) , and set \(x_k=r_kx_0\) , such that
\(d_g(x_0,x_k)=\rho _k.\) 
Put \(y_k=\sigma x_k\) , where \(\sigma =2^m\)  is sufficiently large.
We denote
\(\tau _k=d_g(x_k,y_k).\) 
By (),
\(d_g(x_0,y_k)<3d_g(y_k,\frac{1}{2}y_k):=\sigma _k,\) 
hence
\(\tau _k-\sigma _k\le \rho _k\le \tau _k+\sigma _k.\) 
Since
\(d_g(x,x_0)\le d_g(x,y_k)+d_g(x_0,y_k),\) 
for any \(x\in B_{\tau _k-2\sigma _k}^g(y_k)\) ,
\(d_g(x,x_0)\le \tau _k-2\sigma _k+\sigma _k\le \rho _k.\) 
Similarly, for any \(x\in B_{\rho _k}^g(x_0)\)
\(d_g(x,y_k)\le d_g(x,x_0)+d_g(x_0,y_k)\le \tau _k+2\sigma _k.\) 
Then
\(B^g_{\tau _k-2\sigma _k}(y_k)\subset B^g_{\rho _k}(x_0),\,\,\,\,B^g_{\rho _k}(x_0)\subset B^g_{\tau _k+2\sigma _k}(y_k).\) 
Let \(u_k=u(r_kx)r^\frac{n-2}{2}c_k\) , where \(c_k\)  is as in Lemma REF . Since
\(u_k\)  converges to \(|x|^{2-n}\) , and
\(\frac{\sigma _k}{\tau _k}\rightarrow \frac{3}{\sigma -1},\) 
by choosing \(\sigma \)  sufficiently large, for a fixed \(\epsilon \)  and
sufficiently large \(k\) ,
\(B^g_{(1-\epsilon )\tau _k}(y_k)\subset B^g_{\rho _k}(x_0)\subset B^g_{(1+\epsilon )\tau _k}(y_k).\) 
Next, we estimate
\(\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k),g)}{(\lambda \tau _k)^n}.\) 
By (), () and (), we have
\(\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{{(\lambda \tau _k)}^n}&\le &\frac{\mathrm {vol}(B_{{\lambda \tau _k}}^g(y_k),g)}{{(\lambda \tau _k)}^n}\\&=&\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)+\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\setminus B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)+\mathrm {vol}(B\setminus B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)+C\mathrm {vol}(B_{2\sigma r_k}\setminus B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{(\lambda \tau _k)^n}+C\frac{d_g^n(2\sigma r_kx,\sigma r_kx)}{(\lambda \tau _k)^n}\\&\le &\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{(\lambda \tau _k)^n}+C\frac{\sigma _k^n}{\lambda ^n\tau _k^n}.\) 
It is easy to check that
\(\frac{\mathrm {vol}(B_{\lambda \tau _k}^g(y_k)\cap B_{\sigma r_k},g)}{(\lambda \tau _k)^n}\rightarrow \frac{\mathrm {vol}(B^{g_\infty }_{\lambda (1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma },g_\infty )}{(\lambda (1-\sigma ^{-1}))^n}.\) 
To calculate \(\mathrm {vol}(B^{g_\infty }_{(1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma },g_\infty )\) , we use the coordinates change
\(B\rightarrow \mathbb {R}^n\setminus B: \,\,\,\,x\rightarrow z=\frac{x}{|x|^2}.\) 
In the new coordinates, \(g_\infty =g_{euc}\) , and
\(B^{g_\infty }_{\lambda (1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma }=B_{\lambda (1-\sigma ^{-1})}(\sigma ^{-1}\frac{x_0}{|x_0|})\setminus B_{\sigma ^{-1}}.\) 
Thus
\(\mathrm {vol}(B^{g_\infty }_{\lambda (1-\sigma ^{-1}) }(\sigma x_0/|x_0|)\cap B_{\sigma },g_\infty )=V_n((\lambda (1-\sigma ^{-1}))^n-\sigma ^{-n}).\) 
Then we can select \(\sigma \)  to be sufficiently large, such that
\(1-\epsilon \le \frac{\mathrm {vol}(B_{(1+\epsilon )\tau _k}^g(y_k),g)}{V_n((1+\epsilon )\tau _k)^n},\,\,\,\,\frac{\mathrm {vol}(B_{(1-\epsilon )\tau _k}^g(y_k),g)}{V_n((1-\epsilon )\tau _k)^n}\le (1+\epsilon )\) 
when \(k\)  is sufficiently large. By (REF ) and (REF ),
\(1-C\epsilon \le \frac{\mathrm {vol}(B_{\rho _k}(x_0))}{V_n\rho _k^n}\le 1+C\epsilon \) 
when \(k\)  is sufficiently large.
\(\hfill \Box \)
Corollary 3.5 
Assume \((M,g)\)  is conformal compact with finite \(\Vert R\Vert _{L^\frac{n}{2}}\) .
If \(Ric\ge 0\) , and if \((M,g)\)  is complete and noncompact, then \((M,g)=\mathbb {R}^n\) .

By Bishop-Gromov Theorem, \(\mathrm {vol}(B_r^g(x),g)\le V_nr^n\) . Then by a result in [3]}, there exists a compact manifold \((M_0,g_0)\)  and a finite set \(A\subset M_0\) , such that \((M,g)\)  is conformal to \((M_0\setminus A,g_0)\) .
By Corollary REF , \(A\) 
contains a single point, so the corollary follows from the Bishop-Gromov Theorem.
\(\hfill \Box \)
Next, we derive a stronger version of Lemma REF  and
finish the proof of Theorem REF :
Proposition 3.6 \(w=G^{-1}u\)  is in \(W^{2,p}(B)\)  for any \(p\in [1,\frac{n}{2})\) , where \(G\)  is the Green function defined by
\(-\Delta _{g_0}G=\delta _0,\,\,\,\,G|_{\partial B}=0.\) 

By a direct computation,
\(\Delta _{g_0} u=G\Delta _{g_0} w+2\nabla _{g_0} G\nabla _{g_0} w=-c(n)Ru^\frac{n+2}{n-2}+c(n)R(g_0)u.\) 
Then
\(-\Delta _{g_0} w&=&c(n)G^{-1}Ru^\frac{n+2}{n-2}+2\nabla _{g_0}\log G\nabla _{g_0} w-c(n)R(g_0)w\\&=&c(n)Ru^\frac{4}{n-2}w+2\nabla _{g_0}\log G\nabla _{g_0} w-c(n)R(g_0)w\\&:=&f.\) 
It is well known that \(G=r^{2-n}(1+O(1))\)  near 0. By the gradient estimate
of Cheng and Yau on \(B_{|x|/2}^{g_0}(x)\) ,
\(|\nabla _{g_0}\log G|(x)\le \frac{C}{|x|}\) 
when \(x\)  is small.
First, we show \(w\in L^q\)  for any \(q\) . Indeed, applying () to \(\alpha =q\) , \(\beta =(n-2)q\) ,
we get
\(\int _{B_t}|w|^q\le C\sum _i\int _{B_{2^{-i}t}}|x|^{p(n-2)}|u|^q<C\sum _i((1+\tau )2^{-n})^i<+\infty .\) 
Next, we show \(w\in W^{1,p}(B)\)  for any \(p<n\) . Since
\(|\nabla _{g_0}w|\le G^{-1}|\nabla _{g_0}u|+uG^{-1}|\nabla _{g_0}\log G|\le C(|x|^{n-2}|\nabla _{g_0}u|+|x|^{n-3}u),\) 
we may apply () to \(\alpha =p\)  and \(\beta =(n-3)p\)  to
get
\(\int _{B_t}(|x|^{n-3}|u|)^p<+\infty \) 
and () to \(\alpha =p\)  and \(\beta =(n-2)p\)  to obtain
\(\int _{B_t}|x|^{n-2}|\nabla u|^p<+\infty .\) 
Then \(\int _{B_t}|\nabla w|^p<+\infty \)  for any \(p<n\) . Let \(\varphi \in \mathcal {D}(B)\) , and \(\eta _\epsilon \)  be a cutoff function which is
1 in \(B\setminus B_{2\epsilon }\) , 0 in \(B_\epsilon \) , and satisfies \(|\nabla \eta _\epsilon |<\frac{C}{\epsilon }\) . It is essay to check that \(\eta _\epsilon w\)  is bounded in \(W^{1,p}(B)\) , hence a subsequence of \(\eta _\epsilon w\) 
converges weakly in \(W^{1,p}(B)\) . Obviously, \(w\)  is the limit, hence \(w\in W^{1,p}(B)\) .
Next, we show \(f\in L^p\)  for any \(p<\frac{n}{2}\)  and \(w\)  solves the equation \(-\Delta _{g_0}w=f\)  weakly in \(B\) .
Since
\(|f|\le C(|R|u^\frac{4}{n-2}w+|x|^{-1}|\nabla _{g_0}w|+w),\) 
by the fact that \(|R|u^\frac{4}{n-2}\in L^\frac{n}{2}\) , \(w\in L^q\)  for any \(q>0\)  and \(\nabla w\in L^p\)  for any \(p<n\) ,
one is easy to check that \(f\in L^p\)  for any \(p<\frac{n}{2}\) . Then
\(\int _{B}\nabla _{g_0}\varphi \nabla _{g_0} wdV_{g_0}=\lim _{\epsilon \rightarrow 0}\int _B\nabla _{g_0}\eta _\epsilon \varphi \nabla _{g_0} wdV_{g_0}=\lim _{\epsilon \rightarrow 0}\int _B\eta _\epsilon \varphi f=\int _B\varphi f,\) 
hence \(w\)  is a weak solution.
By the theory of elliptic equations, we can complete the proof
without difficulty.
\(\hfill \Box \)

conformally immersed submanifolds in \(\mathbb {R}^{n+k}\)
In this section, we let \(F:(B\setminus \lbrace 0\rbrace ,g_0) \rightarrow \mathbb {R}^n\)  be a conformal immersion with
\(\Vert A\Vert _{L^n}<+\infty \) , where \(A\)  is the second fundamental form. We set
\(g=F^*(g_0)=u^\frac{4}{n-2}g_0.\) 
Obviously
\(\int _{B}|R|^\frac{n}{2}<+\infty .\) 
By result in [4]}, \((B\setminus \lbrace 0\rbrace ,g)\)  can be extended
to a compact metric space in the case when \(\mathrm {vol}(B\setminus \lbrace 0\rbrace ,g)<+\infty \) . We assume \(\mathrm {vol}(F(B\setminus \lbrace 0\rbrace ))=+\infty \)  in this section. It follows from
Corollary REF  that \((\overline{B}_\frac{1}{2}\setminus \lbrace 0\rbrace ,g)\)  is complete.
The goal of this section is to prove Theorem REF .
Proof of Theorem REF : First of all,
we can find \(c_k\) , such that \(c_k r_k^\frac{n-2}{2} u(r_k x)\) 
converges to \(|x|^{2-n}\)  weakly in \(W^{2,p}_{loc}(\mathbb {R}^n\setminus \lbrace 0\rbrace )\) 
for any \(p<\frac{n}{2}\) .
Set
\(F_k=c_k^\frac{2}{n-2}(F(r_kx)-F(r_kx_0)).\) 
Since
\(\left|\frac{\partial F_k}{\partial x^i}\right|^2=r_k^2c_k^\frac{4}{n-2}(u_k(r_kx))^\frac{4}{n-2}=(c_kr_k^\frac{n-2}{2}u_k(r_kx))^\frac{4}{n-2},\) 
\(|\nabla F_k|\)  is bounded in \(W^{2,p}(B_r\setminus B_\frac{1}{r})\)  by Lemma REF . Thus, we may assume
\(F_k\)  converges weakly in \(W^{3,p}_{loc}\)  to a map \(f_\infty \)  with
\(F_\infty (\mathbb {R}^n)\subset \mathbb {R}^n\)  and
\(\frac{\partial F_\infty }{\partial x^i}\frac{\partial F_\infty }{\partial x^j}=|x|^{-4}\delta _{ij}.\) 
It will be convenient if we change the coordinates
\(x\rightarrow y=\frac{x}{|x|^2}.\) 
In new coordinates,
\(g_\infty (y)=g_{euc}(y).\) 
Then \(F_\infty \)  can be considered as a isometric map from \(\mathbb {R}^n\setminus \lbrace 0\rbrace \) 
to itself.
Let \(\gamma (t)=ty\) , where \(y\in \mathbb {S}^{n-1}\) . Since \(\gamma (t)\)  is a geodesic
in \(\mathbb {R}^n\setminus \lbrace 0\rbrace \) , \(f(\gamma (t))\)  should be a ray of \(\mathbb {R}^n\) .
In addition, it is easy to check that
\(d_{g_\infty }(F_\infty (y^{\prime }),F_\infty (y^{\prime \prime }))\rightarrow 0\)  as \(y^{\prime }\) , \(y^{\prime \prime }\) 
approach 0, then \(\lim \limits _{y\rightarrow 0}f_\infty (y)\)  exists.
Without loss of generality, we assume the limit is 0. Then
\(F_\infty (\gamma (t))=tF_\infty (y)\) .
Since \(F_\infty \)  is conformal, for any \(X\in T\partial B_1=\mathbb {S}^{n-1}\)  with \(|X|=1\) , we have
\(F_{\infty ,*}(X)\bot F_{\infty ,*}(\frac{\partial }{\partial r}),\,\,\,\,|F_{\infty ,*}(X)|=1.\) 
Thus, \(F_\infty |_{ \mathbb {S}^{n-1}}\)  is an isometric map from
\(\mathbb {S}^{n-1}\)  to \(\mathbb {S}^{n-1}\) . Since \(\mathbb {S}^{n-1}\)  is simple-connected, \(F_\infty |_{\mathbb {S}^{n-1}}\)  is a homeomorphism, as follows from the fact that
an isometric map is a covering map. Therefore, we may assume \(F_\infty (y)=y\) .
Consequently, in the origin coordinates, we have
\(F_\infty (x)=\frac{x}{|x|^2}.\) 
Now, we let \(x_k\rightarrow 0\) .
Assume \(a=|x_0|\) , and \(x_k^{\prime }=a\frac{x_k}{|x_k|}\) . Put \(r_k=|x_k|\) , \(y_k=\sigma x_k\) , where \(\sigma =2^{-m}\)  is sufficiently large. Since
\(d_g(\sigma x_k,x_k^{\prime })\le Cd_g(\sigma x_k,2\sigma x_k),\,\,\,\,and \,\,\,\,\frac{d_g(\sigma x_k,2\sigma x_k)}{d_g(x_k,\sigma x_k)}\rightarrow \frac{1}{\sigma },\) 
when \(\sigma \)  is sufficiently large,
\(1-\epsilon \le \frac{d_g(x_k,x_0)}{d_g(x_k,\sigma x_k)}\le 1+\epsilon .\) 
By an arguments as in the proof of Lemma REF , for any \(\tau \in (0,1)\) ,
\(\frac{1-\tau }{2}\le \frac{|F(rx)-F(rx/2)|}{|F(rx/2)-F(rx/4)|}\le \frac{1+\tau }{2}\) 
for any \(x\in \partial B\)  and sufficiently small \(r\) , then for sufficiently large \(k\) ,
\(1-\epsilon \le \frac{|F(x_k)-F(x_0^{\prime })|}{|F(x_k)-F(\sigma x_k)|}<1+\epsilon .\) 
Since
\(\frac{|F(x_k)-F(\sigma x_k)|}{d_g(x_k,\sigma x_k)}=\frac{|F_k(x_k/|x_k|)-F_k(\sigma x_k/|x_k|)|}{d_{g_k}(x_k/|x_k|,\sigma x_k/|x_k|)}\rightarrow 1,\) 
\(1-C\epsilon <\frac{|F(x_k)-F(x_0)|}{d_g(x_k,x_0)}\le 1+C\epsilon \) 
when \(k\)  is sufficiently large.
\(\hfill \Box \)

 4 dimensional Gauss-Bonnet-Chern formulas
In this section, we assume \(n=4\)  and discuss the Gauss-Bonnet-Chern formulas.
For our purpose, we have to estimate \(\int _{B_{t_1}\setminus B_{t_2}}\Delta _gRdV_g\)  and \(\int _{B_{t_1}\setminus B_{t_2}}\Delta _{g_0}^2\log udV_{g_0}\) , or \(\int _{\partial B_t}n_{\partial B_t,g}(R)dS_{g}\)  and
\(\int _{\partial B_t}\frac{\partial \Delta _{g_0}\log u}{\partial r}dS_{g_0}\) .
For simplicity, we put \(dS_{g_0}=\Theta (r,\theta )d\mathbb {S}^3\) , \(\phi =\log u\)  and
\(F_1(r)=\int _{\mathbb {S}^3}R(r,\theta )u^{2}(r,\theta )\Theta (r,\theta )d\mathbb {S}^3,\,\,\,\,H_1=-\int _{S^3}R(r,\theta )\frac{\partial }{\partial r}(u^{2}(r,\theta )\Theta (r,\theta ))d\mathbb {S}^3\) 
\(F_2(r)=\int _{S^3}\Delta _{g_0}\phi (r,\theta )\Theta (r,\theta )d\mathbb {S}^3,\,\,\,\,H_2=-\int _{S^3}\Delta _{g_0}\phi \frac{\partial }{\partial r}\Theta (r,\theta )d\mathbb {S}^3.\) 
Note that
\(\int _{\partial B_r}n_{g,\partial B_r}(R)dS_g=\int _{S^3}u^2\frac{\partial R}{\partial r}\Theta d\mathbb {S}^3=F_1^{\prime }(r)+H_1(r),\) 
and
\(\int _{\partial B_r}\frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}=F_2^{\prime }(r)+H_2(r).\) 
The key point is that we will find \(r_k\rightarrow 0\) , such that \(F_i^{\prime }(r_k)+H_i(r_k)\)  converges.
The following lemma will play a very important role in the following discussion.
Lemma 5.1 
Let \(f\in C^1[\frac{r_0}{4}, 2r_0]\) , \(h\in C^0[\frac{r_0}{4}, 2r_0]\)  and \(b_1\) , \(b_2\)  are constants. Assume
\(\left|\int _{\frac{r_0}{4}}^{\frac{r_0}{2}}fdt-\frac{3}{32}b_1r_0^2\right|+\left|\int _{r_0}^{2r_0}fdt-\frac{3}{2}b_1r_0^2\right|\le ar_0^2,\,\,\,\,\int _{\frac{r_0}{4}}^{2r_0}|h-b_2|<ar_0.\) 
Then there exists \(\xi \in [r_0/4,2r_0]\)  such that
\(|f^{\prime }(\xi )+h(\xi )-b_1-b_2|\le 12a.\) 

Since we can replace \(f\)  with \(f-b_1r\)  and \(h\)  with \(h-b_2\) , it suffices for our aim to prove the case when \(b_1=b_2=0\) .
By The Mean Value Theorem for Integrals, there exists \(\xi _1\in [r_0/4,r_0/2]\)  and \(\xi _2\in [r_0,2r_0]\) , such that
\(\frac{r_0}{4}f(\xi _1)=\int _\frac{r_0}{4}^\frac{r_0}{2}f(t)dt,\,\,\,\,r_0f(\xi _2)=\int _{r_0}^{2r_0}f(t)dt,\) 
which yields that
\(|f(\xi _1)|\le 4ar_0,\,\,\,\,|f(\xi _2)|\le ar_0.\) 
Then
\(\left|\int _{\xi _1}^{\xi _2}(f^{\prime }+h)\right|\le |f(\xi _1)-f(\xi _2)|+\int _{\xi _1}^{\xi _2}|h|\le 5ar_0+\int _{r_0/4}^{2r_0}|h|\le 6ar_0.\) 
Using the Mean Value Theorem for Integrals again, we can find \(\xi \in [\xi _1,\xi _2]\) , such that
\((\xi _2-\xi _1)|f^{\prime }(\xi )+h(\xi )|\le 6a r_0.\) 
Noting that \(r_0/2<\xi _2-\xi _1\) , we complete the proof.
\(\hfill \Box \)
Lemma 5.2 For any sufficiently small \(r\) , we have
\(\left|\int _r^{2r}F_1(t)dt\right|<\alpha (r)r^2, \,\,\,\,\int _r^{2r}|H_1|(t)dt<\alpha (r)r,\) 
and
\(\left|\int _r^{2r}F_2(t)dt+6\omega _3r^2\right|<\alpha (r)r^2, \,\,\,\,\int _r^{2r}|H_2-12\omega _3|(t)dt<\alpha (r)r,\) 
where \(\lim \limits _{r\rightarrow 0}\alpha (r)=0\) , and \(\omega _3=2\pi ^2\)  is the volume of the 3-dimensional sphere.

We have
\(\left|\int _{r}^{2r} F_1(t)\right|& \le & \int _{r}^{2r} \int _{S^3} |R(g)| u^2 \Theta dSdt \\& =& \int _{B_{2r}\setminus B_r}|R(g)| u^2dV_{g_0}\\&\le &\left(\int _{B_{2r} \setminus B_{r}} | R(g) u^2 |^2 dV_{g_0}\right)^{\frac{1}{2}}\left(\int _{B_{2r} \setminus B_{r}} dV_{g_0}\right)^{\frac{1}{2}} \\&\le & C\Vert R\Vert _{L^2(B_{2r},g)}r^2.\) 
and
\(\int _{r}^{2r} |H_1(t)| dt& \le & 2\int _{r}^{2r} \int _{\mathbb {S}^3}|R(g)| u^2 \frac{\partial \log u}{\partial t}\Theta d\mathbb {S}^3dt+ 2\int _{r}^{2r} \int _{\mathbb {S}^3}|R(g)| u^2\frac{\partial \log \Theta (t,\theta )}{\partial t}(\Theta d\mathbb {S}^3dt) \\&\le & C(\Vert R\Vert _{L^2(B_{2r},g)}\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}+\int _{B_{2r}\setminus B_r}|R(g)| u^2\frac{1}{r}dV_{g_0})\\&\le & C(\Vert R\Vert _{L^2(B_{2r},g)}\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}+Cr\Vert R(g)\Vert _{L^2(B_{2r},g)}.\) 
By Lemma REF ,
\(\Vert \nabla \phi \Vert _{L^2(B_{2r}\setminus B_r,g_0)}\le Cr\Vert Ru^2+R(g_0)\Vert _{L^2(B_{2r}\setminus B_r,g_0)}\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r+r^2),\) 
hence
\(\int _r^{2r}|H_1(t)|dt\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r+r^2).\) 
Next, we discuss \(F_2\) . Since
\(-\Delta _{g_0}\phi -|\nabla _{g_0}\phi |^2=c(n)Ru^2-c(n)R(g_0),\) 
we obtain
\(\int _{r}^{2r} F_2(t) dt& =& \int _{B_{2r}\setminus B_r}\Delta _{g_0}\phi dV_{g_0} \\& =& -\int _{B_{2r}\setminus B_r}|\nabla _{g_0} \phi |^2dV_{g_0}-\int _{B_{2r}\setminus B_r}(R(g) u^2-R(g_0))dV_{g_0}.\) 
Note that
\(\int _{B_{2r}\setminus B_r}|R(g) u^2-R(g_0)|dV_{g_0}\le C(\Vert R\Vert _{L^2(B_{2r}\setminus B_r,g)}r^2+r^4).\) 
To get the estimate of \(\int _{r}^{2r}F_2\) , we only need to show that
\(\lim _{r\rightarrow 0}\frac{1}{r^2}\int _{B_{2r}\setminus B_r}|\nabla _{g_0}\phi |^2dV_{g_0}=6\omega _3.\) 
Assume there exists \(r_k\rightarrow 0\) , such that
\(\lim _{k\rightarrow \infty }\frac{1}{r_k^2}\int _{B_{2r_k}\setminus B_{r_k}}|\nabla _{g_0}\phi |^2dV_{g_0}=\lambda \ne 6\omega _3.\) 
Set \(u_k=c_kr_ku(r_kx)\) , where \(c_k\)  is chosen such that \(\int _{\partial B_1}\log u_k=0\) .
By the arguments in Section 3, \(\log u_k(x)\)  converges to
\(-\log |x|^{-2}\)  weakly in \(W^{2,p}_{loc}(\mathbb {R}^4\setminus \lbrace 0\rbrace )\) . Then, after passing to a subsequence,
\(\int _{B_2\setminus B_1}|\nabla _{g_0(r_kx)}\log u_k|^2\)  converges to \(6\omega _3\) , hence
\(\lim _{k\rightarrow +\infty }\frac{1}{r_k^2}\int _{B_{2r_k}\setminus B_{r_k}}|\nabla _{g_0}\phi |^2dV_{g_0}=\lim _{k\rightarrow +\infty }\int _{B_2\setminus B_1}|\nabla _{g_0(r_kx)}\log u_k|^2dV_{g_0(r_kx)}=6\omega _3.\) 
A contradiction.
Lastly, we calculate \(\int _r^{2r}|H_2-12\omega _3|\) :
\(\int _{r}^{2r} |H_2(t) -12\omega _3|dt\le \int _{B_{2r}\setminus B_r}\left||\nabla _{g_0} \phi |^2\frac{\partial \log \Theta }{\partial r}-\frac{12}{\Theta }\right|dV_{g_0}+C\int _{B_{2r}\setminus B_r}|R(g) u^2-R(g_0)|\frac{1}{r}dV_{g_0}.\) 
The same argument as above shows that
\(\lim _{r\rightarrow 0}\frac{1}{r}\int _{B_{2r}\setminus B_r}\left||\nabla _{g_0} \phi |^2\frac{\partial \log \Theta }{\partial r}-\frac{12}{\Theta }\right|dV_{g_0}=0.\) 
This completes the proof.
\(\hfill \Box \)
We will give several applications here. First, we study the integrability of \(\Delta _{g} R\) :
Lemma 5.3 There exists \(r_k \rightarrow 0\)  such that
\(\int _{B_{\frac{1}{2}}\setminus B_{r_k}}\Delta _g R(g) dV_g\rightarrow \int _{\partial B_\frac{1}{2}}\frac{\partial R}{\partial r}dS_g.\) 
Here \(B_r\)  is a ball with respect to \(g_0\) .

We have
\(\int _{B_{\frac{1}{2}}\setminus B_{r}}\Delta _g R(g) dV_g=\int _{\partial B_\frac{1}{2}}n_{g,\partial B_\frac{1}{2}}(R)dS_g-\int _{\partial B_r}n_{g,\partial B_r}(R)dS_g.\) 
Applying Lemma REF  to \(b_1=b_2=0\)  and \(f=F_1\) , \(h=H_1\)  , we deduce this lemma from (REF ).
\(\hfill \Box \)
We recall some basic properties of \(Q\) -curvatures, c.f. [5]}, [6]}. On a 4 dimensional manifold, the Paneitz operator is defined as follows:
\(P_{g_0}\varphi =\Delta _{g_0}^2\varphi +div_{g_0}\left(\frac{2}{3}R_{g_0}\nabla _{g_0}\varphi -2Ric_{g_0}^{ij}\varphi _i\frac{\partial }{\partial x^j}\right).\) 
The \(Q\) -curvature of \(g\)  satisfies the following equations:
\(Q(g)=-\frac{1}{12}\Delta _gR(g)-\frac{1}{4}|Ric(g)|^2+\frac{1}{12}R^2,\) 
\(P_{g_0}\phi +2Q(g_0)=2Q(g)e^{4\phi }.\) 
For simplicity, we define
\(T(\phi )=\frac{1}{3}R_{g_0}\frac{\partial \phi }{\partial r}-Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r}).\) 
Lemma 5.4 
There exists \(r_k\rightarrow 0\) , such that
\(\lim _{k\rightarrow +\infty }\int _{B_\frac{1}{2}\setminus B_{r_k}} Q(g)dV_g=\int _{B_\frac{1}{2}}Q(g_0)dV_{g_0}-4\omega _3+\int _{\partial B_\frac{1}{2}}\left(\frac{1}{2}\frac{\partial \Delta _{g_0}\phi }{\partial r}+T(\phi )\right).\) 

We have
\(\int _{B_\frac{1}{2}\setminus B_r}Q_gdV_g&=&\int _{B_\frac{1}{2}\setminus B_r}Q_{g_0}dV_{g_0}+\frac{1}{2}\int _{B_\frac{1}{2}\setminus B_r}P_{g_0}(\phi )dV_{g_0}\\&=&\int _{B_\frac{1}{2}\setminus B_r}Q_{g_0}dV_{g_0}\\&&+\frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\left(\frac{\partial \Delta _{g_0}\phi }{\partial r}+\frac{2}{3}R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}.\) 
First, we check that
\(\int _{\partial B_r}\left(\frac{2}{3}R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}\rightarrow 0.\) 
By (),
\(\lim _{r\rightarrow 0}\int _{\partial B_r}|\nabla _{g_0}\phi |dS_{g_0}= 0.\) 
Then (REF ) follows.
Then, by applying Lemma REF  to \(f=F_2/2\) , \(h=H_2/2\)  and \((b_1,b_2)=(-2\omega _3,6\omega _3)\) , we deduce from (REF )
that there exists \(r_k\) , such that
\(\int _{\partial B_{r_k}}\frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\rightarrow 8\omega _3.\) 
Therefore, we complete the proof.
\(\hfill \Box \)
Next, we discuss the relationship between \(\Vert R\Vert _{L^2}\) 
and \(\Vert Riem\Vert _{L^2}\) :
Lemma 5.5 
We have
\(\int _{B_\frac{1}{2}}|Riem(g)|^2dV_g<+\infty .\) 

It is well-known that
\(Riem(g)=W(g)+\frac{1}{2}(Ric(g)-\frac{1}{6}R(g)g){~\wedge \!\!\!\!\!\bigcirc ~}g,\) 
where \(W\)  is the Weyl tensor and \({~\wedge \!\!\!\!\!\bigcirc ~}\)  is the Kulkarni-Nomizu product. Since \(|W|^2dV_g\)  is conformal invariant, we only need to check
\(Ric(g)\in L^2\)  here.
Recall that
\(Q(g)=-\frac{1}{12}\Delta _gR(g)-\frac{1}{4}|Ric(g)|^2+\frac{1}{12}R^2\) ,
which means that
\(\int _{B_\frac{1}{2}\setminus B_r}|Ric(g)|^2dV_g&=&\frac{1}{3}\int _{B_\frac{1}{2}\setminus B_r}R^2dV_{g}-\frac{1}{3}\int _{B_\frac{1}{2}\setminus B_r}\Delta _gR(g)dV_{g}-4\int _{B_\frac{1}{2}\setminus B_r}Q(g)dV_{g}\\&=&\frac{1}{3}\int _{B_\frac{1}{2}}R^2dV_{g}-4\int _{B_\frac{1}{2}}Q(g_0)dV_{g_0}\\&&-\frac{1}{3}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\frac{\partial R}{\partial r}dS_g-2\int _{\partial (B_\frac{1}{2}\setminus B_r)}(\frac{\partial \Delta _{g_0}\phi }{\partial r}+2T(\phi ))dS_{g_0}\) 
Applying Lemma REF  to \(f=\frac{1}{3}F_1+2F_2\) , \(h=\frac{1}{3}H_1+2H_2\) , \((b_1,b_2)=(-8\omega _3,24\omega _3)\) , we can find \(r_k\rightarrow 0\) , such that
\(\frac{1}{3}\int _{\partial B_{r_k}}\frac{\partial R}{\partial r}dS_g+2\int _{\partial B_{r_k}}(\frac{\partial \Delta _{g_0}\phi }{\partial r}+2T(\phi ))\rightarrow 16\omega _3.\) 
 \(\hfill \Box \)
Lastly, we consider the formula for Pfaffian form:
Lemma 5.6 
We have
\(\int _{B_\frac{1}{2}}Pf(g)&=&-4\omega _3+\int _{B_\frac{1}{2}}Pf(g_0)+\frac{1}{2}\int _{\partial B_\frac{1}{2}} \frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\\&&+\frac{1}{12}\int _{\partial B_\frac{1}{2}}u^2\frac{\partial R}{\partial r}dS_{g_0}-\frac{1}{12}\int _{B_\frac{1}{2}}\Delta _{g_0}R(g_0)dV_{g_0}\\&&+\int _{\partial B_\frac{1}{2}}T(\phi )dS_{g_0}.\) 

Recall that (c.f. [6]})
\(Pf(g)=\frac{1}{8}|W(g)|^2+\frac{1}{12}R^2-\frac{1}{4}|Ric(g)|^2,\) 
where \(W\)  is the Weyl tensor. Since
\(\int _{B}|W(g)|^2dV_{g}=\int _B|W(g_0)|^2dV_{g_0}<+\infty ,\) 
\(Pf(g)\)  is integrable. Since
\(\begin{aligned}Pf(g) & = \frac{1}{8} |W(g)|^2 dV_g+ Q(g) dV_g+ \frac{1}{12} \Delta _g R(g) dV_g, \\Pf(g_0) & = \frac{1}{8} |W(g_0)|^2 dV_{g_0}+ Q(g_0) dV_{g_0}+ \frac{1}{12} \Delta _{g_0} R(g_0) dV_{g_0},\end{aligned}\) 
and
\(P_{g_0} \phi + 2Q(g_0) =2 Q(g) e^{4 \phi },\) 
where \(g = u^2 g_0 = e^{2 \phi } g_0\) ,
we have
\(\int _{B_\frac{1}{2}\setminus B_r} Pf(g)& =& \int _{B_\frac{1}{2}\setminus B_r} Pf(g_0)+ \frac{1}{2}\int _{B_\frac{1}{2}\setminus B_r} P_{g_0} \phi dV_{g_0}\\&& + \frac{1}{12} \int _{B_\frac{1}{2}\setminus B_r} \Delta _g R(g) dV_g-\frac{1}{12} \int _{B_\frac{1}{2}\setminus B_r} \Delta _{g_0} R(g_0) dV_{g_0}\\& =&\int _{B_\frac{1}{2}\setminus B_r}Pf(g_0)+ \frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)} \frac{\partial \Delta _{g_0}\phi }{\partial r}dS_{g_0}\\&&+\frac{1}{12}\int _{\partial (B_\frac{1}{2}\setminus B_r)}u^2\frac{\partial R}{\partial r}dS_{g_0}-\frac{1}{12}\int _{B_\frac{1}{2}\setminus B_r}\Delta _{g_0}R(g_0)dV_{g_0}\\&&+\frac{1}{2}\int _{\partial (B_\frac{1}{2}\setminus B_r)}\left(R_{g_0}\frac{\partial \phi }{\partial r}-2Ric_{g_0}(\nabla _{g_0}\phi ,\frac{\partial }{\partial r})\right)dS_{g_0}.\) 
Apply Lemma REF  to \(f=\frac{1}{12}F_1+\frac{1}{2}F_2\) ,
and \(h=\frac{1}{12}H_1+\frac{1}{2}H_2\) , \((b_1,b_2)=(-2\omega _3,6\omega _3)\) , which suffices to complete the proof.
\(\hfill \Box \)
Theorem REF , REF  and REF  can be deduced from
Lemma REF , REF  and REF  easily.
